Let f(x) = 2x³ – 1822 – 240x. Complete this problem without a graphing calculator. (a) The derivative of f(x) is f'(x) = (b) As a comma-separated list, the critical points of ƒ are x = Since f is continuous on the closed interval [-5, 18], f has both an absolute maximum and an absolute minimum on the interval [-5, 18] according to the Extreme Value Theorem. To find the extreme values, we evaluate f at the endpoints and at the critical points. (c) As a comma-separated list, the y-values corresponding to the critical points and endpoints are y =

Let f(x) = 2x³ – 1822 – 240x. Complete this problem without a graphing calculator. (a) The derivative of f(x) is f'(x) = (b) As a comma-separated list, the critical points of ƒ are x = Since f is continuous on the closed interval [-5, 18], f has both an absolute maximum and an absolute minimum on the interval [-5, 18] according to the Extreme Value Theorem. To find the extreme values, we evaluate f at the endpoints and at the critical points. (c) As a comma-separated list, the y-values corresponding to the critical points and endpoints are y =

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter6: Applications Of The Derivative

Section6.CR: Chapter 6 Review

Problem 1CR

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